## 4I.9

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Karina Koo 2H
Posts: 49
Joined: Fri Sep 28, 2018 12:24 am

### 4I.9

9.47 Initially a sample of ideal gas at 323 K occupies 1.67 L at 4.95 atm. The gas is allowed to expand to 7.33 L by two pathways: (a) isothermal, reversible expansion; (b) isothermal, irreversible free expansion. Calculate Stot, S, and Ssurr for each pathway

I dont understand how to solve part B. according to the student manual, change in U = 0. How do we know this? Also, how would we know that because of that, the change of S for surr is 0 as well?

Tam To 1B
Posts: 72
Joined: Fri Sep 28, 2018 12:25 am

### Re: 4I.9

Remember that in a reversible reaction, any small change in a variable, such as pressure, of the system will cause a similar change in the surroundings so that there is essentially no change in internal energy. ( U = 0 )
Because in part b the pathway is irreversible, this means that any change in variable doesn't reverse the process, so the change of entropy of the surroundings = 0. There is only change happening in the system so the change in the total entropy is the same as the change in entropy of the system.

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

### Re: 4I.9

When calculating S for either reaction, you will use the equation delta S = q(rev) / T. Even though the reaction in b. is not reversible, since entropy is a state function, as long as the initial and final states are the same in both reactions, you can just use that same equation. As for the change in entropy for the surroundings, for part b. this is because it is free expansion (expansion against a vacuum). This does no work because external pressure is 0. If external pressure is 0, work is 0, meaning q(surr) is also 0, making delta S = 0. Hope this helps.